Question

I need to generate such a 32-bit binary sequence: it has eight ‘1’ and the rest bits is ‘0’. So written in hex, some of the sequences will be like these:

000000FF
000001FE
000001FD
000001FB
000001F7
...

The problem is to choose 8 from 32, so there are 10518300 combinations. In other words, there are 10518300 sequences like my examples. I will appreciate it if you give me any suggestions or algorithm to generate my desired sequences.

Answer 1

This problem is about making permutations of zeros and ones. The easiest to code solution is to use next_permutation and a vector<bool>.

Prepare a vector with the earliest permutation in lexicographic order (ones are at the back). Run next_permutation until it returns false. Here is a demo code that generates all 8-bit sequences with three bits set:

vector<bool> data(8, false);
data[7] = data[6] = data[5] = true;
do {
    for (int i = 0 ; i != data.size() ; i++) {
        cout << (int)data[i];
    }
    cout << endl;
} while (next_permutation(data.begin(), data.end()));

Here is a link to ideone with a running demo.

Your program will need a 32-bit vector with the last eight elements set to 1. Instead of printing the elements of the sequence, you would need to convert them to a 32-bit int, and store them in the output container.